We are given that the concentration of NaOH is 0.0003 M and are asked to calculate the pH
We know that NaOH dissociates by the following reaction:
NaOH → Na⁺ + OH⁻
Which means that one mole of NaOH produces one mole of OH⁻ ion, which is what we care about since the pH is affected only by the concentration of H⁺ and OH⁻ ions
Now that we know that one mole of NaOH produces one mole of OH⁻, 0.0003M NaOH will produce 0.0003M OH⁻
Concentration of OH⁻ (also written as [OH⁻]) = 3 * 10⁻⁴
<u>pOH of the solution:</u>
pOH = -log[OH⁻] = -log(3 * 10⁻⁴)
pOH = -0.477 + 4
pOH = 3.523
<u>pH of the solution:</u>
We know that the sum of pH and pOH of a solution is 14
pH + pOH = 14
pH + 3.523 = 14 [subtracting 3.523 from both sides]
pH = 10.477
Answer:
The formula for water is . The oxidation number of hydrogen is +1. Since there are two of them, the hydrogen atoms contribute to a charge of +2. The water molecule is neutral; therefore, the oxygen must have an oxidation number of to balance the charge.
Answer:
55.75g
Explanation:
From
m/M = CV
Where
m= required mass of solute
M= molar mass of solute
C= concentration of solution
V= volume of solution=675ml
Molar mass of solute= 3(23) + 31 + 4(16)= 69+31+64=164gmol-1
Number of moles of sodium ions present= 1.5× 675/1000= 1.01 moles
Since 1 mole of Na3PO4 contains 3 moles of Na+
It implies that 1.01/3 moles of Na3PO4 are present in solution= 0.34moles
mass of Na3PO4= number of moles × molar mass= 0.34 × 164 =55.75g
Answer:
12 oxygen atoms are in 4 molecules of HNO3?
Explanation:
the amounts of atoms of all the component in HNO3, which are 1 atom of Hydrogen, 1 atom of Nitrogen and 3 atoms of Oxygen.
Answer: The value of equilibrium constant for new reaction is 
Explanation:
The given chemical equation follows:
The equilibrium constant for the above equation is 
We need to calculate the equilibrium constant for the equation of 3 times of the above chemical equation, which is:
The equilibrium constant for this reaction will be the cube of the initial reaction.
If the equation is multiplied by a factor of '3', the equilibrium constant of the new reaction will be the cube of the equilibrium constant of initial reaction.
The value of equilibrium constant for reverse reaction is:
Hence, the value of equilibrium constant for new reaction is
